Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(\displaystyle\lim_{x \to 0}\left(\dfrac{\sin 3x}{x^3} + \dfrac{a}{x^2} + b\right) = 0\), then the value of \((a+b)\) equals:</p>
<p>(a) 0</p>
<p>(b) \(\dfrac{1}{2}\)</p>
<p>(c) \(\dfrac{3}{2}\)</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: For the limit to exist and equal 0, the coefficients of negative powers of x must vanish individually when we expand sin(3x) using Taylor series. The expansion sin(3x) = 3x - (3x)³/6 + ... gives 3x/x³ - 9x³/(6x³) + ... = 3/x² - 3/2 + ..., so we need a = -3 and the constant term to sum to 0.
<p><strong>Step 1:</strong> Expand sin(3x) using Taylor series: sin(3x) = 3x - (3x)³/3! + (3x)⁵/5! - ... = 3x - 9x³/2 + 81x⁵/40 - ...</p><p><strong>Step 2:</strong> Rewrite the limit expression by substituting the expansion:</p><p>lim[x→0](sin(3x)/x³ + a/x² + b) = lim[x→0](3x/x³ - 9x³/(2x³) + ... + a/x² + b)</p><p>= lim[x→0](3/x² - 9/2 + ... + a/x² + b)</p><p><strong>Step 3:</strong> Combine like powers of x:</p><p>= lim[x→0]((3+a)/x² + (-9/2 + b) + higher order terms)</p><p><strong>Step 4:</strong> For this limit to exist and equal 0, the coefficient of 1/x² must be zero, and the constant term must be zero:</p><p>3 + a = 0 ⟹ a = -3</p><p>-9/2 + b = 0 ⟹ b = 9/2</p><p><strong>Step 5:</strong> Calculate a + b = -3 + 9/2 = -6/2 + 9/2 = 3/2</p><p>∴ Answer: C (a + b = 3/2)</p>
Correct Answer: C